Hi,
I have been trying everything I can to solve this, im sure its not that hard ,but I am just not seeing it. Well here goes:

A fly wheel of radius 0.4m, rotates with an angular displacement (theta) given by the expression:

(theta)= 4t^2 - 3t +2

Where theta is in radians and t is in seconds.

a) What is the tangential accleration of ap oin on the outside rim of the wheel?

I think At= ra
so At= (.4m)( 8 rad/s)= 3.2 rad/s?

What is the centripetal acceleration of a point on the outside rim of the wheel?

I dont know how to go about this I know Ac= V^2/r. I am not sure how to get velocity out of this though I know velocity is the 2nd derivative ,but I dont get how I would plug that in?

c) What is the total acceleration of a point on the outside rim of the wheel?

Would this just be tangential accelration plus centripetal accelration?

Any help would be nice, im really trying I just do not understand this problem.

a) maybe (4t^2-3t+2)'=8t-3; (8t-3)'=8 or maybe tan(8) acceleration. Which curse excersise it is?

Part (a) is correct, although the units should be in m/s^2. You can get that by converting the angular acceleration (8 rad/s^2) into m/s^2 with a converting factor of 1 rad = 0.4m (for this problem).

(B) You're going to use the Ac= V^2/r equation that you mentioned. The thing that might be tripping you up is that V is a function of time, so centripetal acceleration will also be a function of time. By the way, velocity is the first derivative of position, not the second.

© Yes, the total acceleration is the sum of the tangential and centripetal accelerations. However, because the accelerations are vectors at right angles to each other, you'll have to add them using the Pythagorean theorem in order to get the magnitude. You can use some trigonometry to find the direction of the total acceleration. Draw pictures to help you visualize what's going on.

If you're 'not seeing it' then at least tell your teacher - it is possible they're not teaching it properly and they need to know that. FWIW there's a lot going on there and I can't see either.